将变量传递给从命令行运行的php脚本 [英] Pass variable to php script running from command line

问题描述

我有一个php文件，需要从命令行运行（通过crontab）。但我需要传递'type = daily'到文件，但我不知道how.i尝试

  php myfile。 php？type = daily 
  

但此错误是结果：

无法打开输入文件：myfile.php？type = daily

$ <$ p $ <$>

解决方案

在 $ _ GET 数组中）仅对Web访问的页有效。

您需要像 php myfile.php daily 一样调用它，并从<$ c $ argv [0] <$ cv> $ argv code>将 myfile.php ）。

如果网页也用作网页，有两个选项，你可以考虑。使用shell脚本和wget访问它，并从cron调用它：

 ＃！/ bin / sh 
 http://location.to/myfile.php?type=daily 
  

或检入PHP档案是否从命令行调用：

  if（defined（'STDIN'））{
 $ type = $ argv [1]; 
} else {
 $ type = $ _GET ['type']; 
} 
  

（注意：您可能需要/想检查 $ argv 实际上包含足够的变量，如

I have a php file that is needed to be run from command line (via crontab). but i need to pass 'type=daily' to the file but i don't know how.i tried

php myfile.php?type=daily

but this error was result:

Could not open input file: myfile.php?type=daily

what can i do?

解决方案

The ?type=daily argument (ending up in the $_GET array) is only valid for web-accessed pages.

You'll need to call it like php myfile.php daily and retrieve that argument from the $argv array (which would be $argv[1], since $argv[0] would be myfile.php).

If the page is used as a webpage as well, there are two options you could consider. Either accessing it with a shell script and wget and call that from cron:

#!/bin/sh
wget http://location.to/myfile.php?type=daily

Or check in the php file whether it's called from the commandline or not:

if (defined('STDIN')) {
  $type = $argv[1];
} else { 
  $type = $_GET['type'];
}

(Note: You'll probably need/want to check if $argv actually contains enough variables and such)

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